Simplify; express your answer in exponential form. Assume $t\neq 0, r\neq 0$. $\dfrac{{(t^{3})^{5}}}{{(t^{-1}r^{2})^{-4}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${t^{3}}$ to the exponent ${5}$ . Now ${3 \times 5 = 15}$ , so ${(t^{3})^{5} = t^{15}}$ In the denominator, we can use the distributive property of exponents. ${(t^{-1}r^{2})^{-4} = (t^{-1})^{-4}(r^{2})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(t^{3})^{5}}}{{(t^{-1}r^{2})^{-4}}} = \dfrac{{t^{15}}}{{t^{4}r^{-8}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{15}}}{{t^{4}r^{-8}}} = \dfrac{{t^{15}}}{{t^{4}}} \cdot \dfrac{{1}}{{r^{-8}}} = t^{{15} - {4}} \cdot r^{- {(-8)}} = t^{11}r^{8}$.